View of On Topological Pseudo-UP Algebras Based on Pseudo-UP Ideals

Prof. Alias B. Khalaf

, Mahmoud A.Yousef

Department of Mathematics, College of Science, University of Duhok, Kurdistan Region, Iraq.Email: aliasbkhalaf@uod.ac b

Department of Mathematics, College of Basic Education, University of Duhok, Kurdistan Region, Iraq. Email: mahmood.a.y.91@gmail.com

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Abstract: The aim of this paper is to introduce the concept of pseudo-UP ideals defined on a pseudo-UP algebra and by usingthis concept we study a uniform structure on a pseudo-UP algebra. Also, we study some properties of the topology whichis generated by a filter base on pseudo-UP algebra. Moreover, several results are obtained using the concept of pseudo-UPhomomorphisms.

Keywords:UP-algebra; topological UP-algebra; pseudo-UP algebra; topological pseudo-UP algebra; pseudo-UP

homomorphism.

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I. Introduction

Recently, studying and investigating the topological properties of several types of algebras has becomeof interest to many researchers.The topological concepts of (BCK/BCC/ BE)- algebras are given in [4,1,5]. In 1998, Lee and Ryu investigated topological BCK-algebras and determined several topological featuresof this structure.In 2008, Ahn and Kwon introduced the concept of topological BCC-algebras. In 2017, Mehrshad and Golzarpoorinvestigated some characteristics of the topological BE-algebras and uniformBE-algebras. In this same year, Iampan[2]presented a new class of algebras called UP-algebras whichis an extension of KU-algebras [6] introduced by Prabpayak and Leerawat in 2009. Later in 2019, Satirad and Iampan[10]presented and established further characteristics of the topological UP-algebras.In 2020, Romano [7] introduced another class of algebras called pseudo-UP algebras as an extension of UP- algebras. Also, he studied the concepts of pseudo-UP filers and pseudo-UP ideals of pseudo-UP algebras in [8]. Furthermore, he introduced the concept of homomorphisms between pseudo-UP algebras in [9].

This paper is formatted as follows:In Section 2, we present some definitions and propositions on pseudo-UP algebras and topologies which are needed to develop this paper. In Section 3, we study the congruence relation on pseudo-UP ideals. In Section 4, we study the uniform topology on pseudo-UP algebra.We employ the congruence relationship for the uniform topology to create uniform structures based onpseudo-UP ideals of pseudo-UP algebras. Also, we show that topological pseudo-UP algebra is pseudo-UP algebra with uniform topology. Additionally, several characteristics are acquired. In Section 5, weintroduce the filter base on pseudo -UP algebra to generate a topology on pseudo-UP algebra.

2. Preliminaries

In this section, we provide some basic information and observations on pseudo-UP algebras andtopological concepts which are essential for this paper.

Definition 2.1.[7]A pseudo-UP algebra is a structure ((𝑋,≤), ∙ , ∗, 0) where ≤ is a binary operation on a set 𝑋, ∙

and ∗ are two binary operations on 𝑋 if 𝑋 satisfies the following axioms: for all 𝑥, 𝑦, 𝑧 ∈ 𝑋, 1. (𝑦 ∙ 𝑧) ≤ (𝑥 ∙ 𝑦) ∗ (𝑥 ∙ 𝑧) and (𝑦 ∗ 𝑧) ≤ (𝑥 ∗ 𝑦) ∙ (𝑥 ∗ 𝑧).

2. 𝑥 ≤ 𝑦 and 𝑦 ≤ 𝑥 then 𝑥 = 𝑦. 3. (𝑦 ∙ 0) ∗ 𝑥 = 𝑥 and (𝑦 ∗ 0) ∙ 𝑥 = 𝑥.

4. 𝑥 ≤ 𝑦 ⇔ 𝑥 ∙ 𝑦 = 0 and 𝑥 ≤ 𝑦 ⇔ 𝑥 ∗ 𝑦 = 0.

Proposition 2.2. [7] In a pseudo-UP algebra ((𝑋, ≤),∙, ∗, 0) the following statements hold: for all 𝑥 ∈ 𝑋,

1. 𝑥 ∙ 0 = 0 and 𝑥 ∗ 0 = 0, 2. 0 ∙ 𝑥 = 𝑥 and 0 ∗ 𝑥 = 𝑥, and

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3. 𝑥 ∙ 𝑥 = 0 and 𝑥 ∗ 𝑥 = 0.

Proposition 2.3. [7] In a pseudo-UP algebra ((𝑋, ≤), ∙, ∗, 0) the following statements hold: for all 𝑥,𝑦 ∈ 𝑋,

1. 𝑥 ≤ 𝑦 ∙ 𝑥, and 2. 𝑥 ≤ 𝑦 ∗ 𝑥.

Definition 2.4. [8] A non-empty subset ℐ of a pseudo-UP algebra 𝑋 is said to be a pseudo-UP ideal of 𝑋 if it

satisfies: for all 𝑥, 𝑦 ∈ 𝑋, 1. 0 ∈ ℐ,

2. 𝑥 ∙ (𝑥 ∗ 𝑧) ∈ ℐ and 𝑦 ∈ ℐ then 𝑥 ∙ 𝑧 ∈ ℐ, and 3. 𝑥 ∗ (𝑥 ∙ 𝑧) ∈ ℐ and 𝑦 ∈ ℐ then 𝑥 ∗ 𝑧 ∈ ℐ.

Proposition 2.5. [8] Let ℐ be a pseudo-UP ideal of a pseudo-UP algebra 𝑋, then the following statements hold: for

all 𝑥,𝑦 ∈ 𝑋,

1. if 𝑥 ∈ ℐ and 𝑥 ∙ 𝑦 ∈ ℐ then 𝑦 ∈ ℐ, and 2. if 𝑥 ∈ ℐ and 𝑥 ∗ 𝑦 ∈ ℐ then 𝑦 ∈ ℐ.

Definition 2.6. [8] A non-empty subset ℱ of a pseudo-UP algebra 𝑋 is said to be a pseudo-UP filter of 𝑋 if it

satisfies: for all 𝑥, 𝑦 ∈ 𝑋, 1. 0 ∈ ℱ,

2. 𝑥 ∙ 𝑦 ∈ ℱ and 𝑥 ∈ ℱ then 𝑦 ∈ ℱ, and 3. 𝑥 ∗ 𝑦 ∈ ℱ and 𝑥 ∈ ℱ then 𝑦 ∈ ℱ.

Definition 2.7. [9] Let ((𝑋,≤), ∙, ∗, 0) and ((𝑌,≤_𝑌), ∙_𝑌, ∗_𝑌, 0_𝑌)be two pseudo-UP algebras. A map 𝑓: 𝑋 → 𝑌

is a pseudo-UP homomorphism if

𝑓(𝑥 ∙ 𝑦) = 𝑓(𝑥) ∙𝑌𝑓(𝑦) and𝑓(𝑥 ∙ 𝑦) = 𝑓(𝑥) ∗𝑌𝑓(𝑦),

for all 𝑥, 𝑦 ∈ 𝑋. Moreover, 𝑓 is a pseudo-UP isomorphism if it is bijective.

Definition 2.8. A pseudo-UP algebra 𝑋 is said to be negative implicative if it satisfies the condition: for all

𝑥,𝑦, 𝑧 ∈ 𝑋,

(𝑥 ∙ 𝑦) ∙ (𝑥 ∙ 𝑧) = 𝑥 ∙ (𝑦 ∙ 𝑧) 𝑎𝑛𝑑 (𝑥 ∗ 𝑦) ∗ (𝑥 ∗ 𝑦) = 𝑥 ∗ (𝑦 ∗ 𝑧).

Example 2.9. Let 𝑋 = {0,𝑎, 𝑏, 𝑐} and the two binary operations ∙ and ∗ defined by the following Cayley tables:

Table1. A negative implicative pseudo-UP algebra

Then it isclear that ((𝑋,≤), ∙, ∗,0) is a pesudo-UP algebra and satisfies the negative implicative condition. In the remainder of this section, we introduce some topological concepts, by (𝑋,𝜏) or 𝑋 we mean a topological space. Let 𝐴 be a subset 𝑋, the closure of 𝐴 is defined by𝑐𝑙(𝐴) = {𝑥 ∈ 𝑋: ∀ 𝑂 ∈ 𝜏 such that 𝑥 ∈ 𝑂,𝑂 ∩ 𝐴 ≠ 𝜙}. The set of all interior points of 𝐴 denoted by 𝑖𝑛𝑡(𝐴) and defined as 𝑖𝑛𝑡(𝐴) = ⋃{𝑈: 𝑈 ∈ 𝜏 and 𝑈 ⊆ 𝐴}. Let 𝑓: (𝑋,𝜏) → (𝑌,𝜏𝑌) be a function, then 𝑓 is continuous if the opposite image of each open set in 𝑌 is open 𝑋. Also,

𝑓 is called closed (resp., open) map if the image of each closed (resp., open) set in 𝑋 is closed (resp., open) set in 𝑌. Furthermore, 𝑓 is homeomorphism if 𝑓 is isomorphism, continuous, and open. All topological concepts above can be found in all texts of general topology.

Definition 2.10. [11] A pseudo-UP algebra ((𝑋, ≤), ∙, ∗,0) with a topology 𝜏 is called a topological pesudo-UP

algebra (for short TPUP-algebra) if for each open set 𝑂 containing 𝑥 ∙ 𝑦 and for each open set 𝑊 containing 𝑥 ∗ 𝑦, there exist open sets 𝑈₁ and 𝑉₁ (𝑈₂ and 𝑉₂) containing 𝑥 and 𝑦 respectively such that 𝑈₁∙ 𝑉₁⊆ 𝑂 and 𝑈₂∗ 𝑉₂⊆ 𝑊 for all 𝑥,𝑦 ∈ 𝑋. c b a 0 ∙ c b a 0 0 c b 0 0 a c 0 a 0 b 0 b a 0 c c b a 0 ∗ c b a 0 0 c b 0 0 a c 0 0 0 b 0 b a 0 c

Proposition 2.11. [11] Let ((𝑋, ≤), ∙, ∗, 0, 𝜏) be a TPUP-algebra and 𝑀₀ be the minimal open set containing 0. If 𝑥 ∈ 𝑀₀ then 𝑀₀ is the minimal open set containing 𝑥.

Proposition 2.11. [11] Let ((𝑋, ≤), ∙, ∗, 0, 𝜏) be a TPUP-algebra and 𝑀_𝑥, 𝑀_𝑦be two minimal open sets containing 𝑥, 𝑦 respectively. If 𝑥 ∙ 𝑦, 𝑥 ∗ 𝑦 ∉ 𝑀₀, then 𝑦 ∉ 𝑀_𝑥 and 𝑥 ∉ 𝑀_𝑦 where 𝑥 ≠ 0 and 𝑦 ≠ 0.

3. On pseudo-UP ideals

In this section, we give some properties of pseudo-UP ideals of pseudo-UP algebras.

Proposition 3.1. Let 𝑋 be a pseudo-UP algebra and {ℐ_𝑖}_𝑖∈𝐴 be a family of pseudo-UP ideals of 𝑋. Then ∩_𝑖∈𝐴ℐ_𝑖 is a UP-ideal of 𝑋.

Proof. Clearly, if 0 ∈ ℐ_𝑖 for all 𝑖 ∈ 𝐴, then 0 ∈∩_𝑖∈𝐴ℐ_𝑖. Let 𝑥, 𝑦, 𝑧 ∈ 𝑋 such that 𝑥 ∙ (𝑦 ∗ 𝑧) ∈ ∩_𝑖∈𝐴ℐ_𝑖,𝑥 ∗ (𝑦 ∙ 𝑧) ∈ ∩𝑖∈𝐴ℐ𝑖 and 𝑦 ∈ ∩𝑖∈𝐴ℐ𝑖. Hence, 𝑥 ∙ (𝑦 ∗ 𝑧) ∈ ℐ𝑖, 𝑥 ∗ (𝑦 ∙ 𝑧) ∈ ℐ𝑖 and 𝑦 ∈ ℐ𝑖 for all 𝑖 ∈ 𝐴. Since ℐ𝑖is a pseudo-UP

ideal of 𝑋, then 𝑥 ∙ 𝑧 ∈ ℐ_𝑖and 𝑥 ∗ 𝑧 ∈ ℐ_𝑖for all 𝑖 ∈ 𝐴. Therefore, 𝑥 ∙ 𝑧 ∈∩_𝑖∈𝐴ℐ_𝑖and 𝑥 ∗ 𝑧 ∈∩_𝑖∈𝐴ℐ_𝑖Hence,∩_𝑖∈𝐴ℐ_𝑖is a pseudo-UP ideal of 𝑋.

Definition 3.2.Let ℐ be a pseudo-UP ideal of a pseudo-UP algebra 𝑋. Define the relation ∼_ℐon 𝑋 as follows: for all 𝑥,𝑦 ∈ 𝑋,

𝑥 ∼ℐ𝑦 if and only if 𝑥 ∙ 𝑦, 𝑥 ∗ 𝑦 ∈ ℐand 𝑦 ∙ 𝑥, 𝑦 ∗ 𝑥 ∈ ℐ.

Example 3.3. Let 𝑋 = {0, 𝑎, 𝑏, 𝑐} with two binary operations ∙ and ∗ defined by the following Cayley tables:

Table2. A psudo-UP algebra

By easy calculation, we can check that((𝑋,≤), ∙, ∗, 0) is a pseudo-UP algebra and ℐ = {0,𝑎, 𝑐} is a pseudo-UP ideal of a pseudo-UP algebra 𝑋. Then

∼ℐ= {(0,0),(0,a), (a,0),(0,c), (c,0),(a, c),(c, a),(a, a), b, b),(c, c)},

so, we can see that ∼_ℐ is an equivalence relation on 𝑋.

Definition 3.4.An equivalence relation 𝑅 on a pseudo-UP algebra 𝑋 is called a congruence relation if for all

𝑥,𝑦, 𝑢,𝑣 ∈ 𝑋,

𝑥 𝑅 𝑢 and 𝑦 𝑅 𝑣 implies 𝑥 ∙ 𝑦 𝑅 𝑢 ∙ 𝑣 and 𝑥 ∗ 𝑦 𝑅 𝑢 ∗ 𝑣.

Proposition 3.5.If ℐ is a pseudo-UP ideal of a pseudo-UP algebra 𝑋, then the binary relation ∼_ℐ defined as in Definition 3.4, is a congruence relation on 𝑋.

Proof.Reflexive: For all 𝑥 ∈ 𝑋, 𝑥 ∙ 𝑥 = 0 and 𝑥 ∗ 𝑥 = 0. Since ℐ is a pseudo-UP ideal of 𝑋, then 𝑥 ∙ 𝑥 = 0 ∈ ℐ and

𝑥 ∗ 𝑥 = 0 ∈ ℐ. Therefore, 𝑥 ∼ℐ𝑥.

Symmetric: Let 𝑥,𝑦 ∈ 𝑋 such that 𝑥 ∼_ℐ𝑦. Then we have 𝑥 ∙ 𝑦, 𝑥 ∗ 𝑦 ∈ ℐ, and 𝑦 ∙ 𝑥, 𝑦 ∗ 𝑥 ∈ ℐ , so 𝑦 ∙ 𝑥, 𝑦 ∗ 𝑥 ∈ ℐ and 𝑥 ∙ 𝑦, 𝑥 ∗ 𝑦 ∈ ℐ. Therefore, 𝑦 ∼ℐ 𝑥.

Transitive: Let 𝑥, 𝑦, 𝑧 ∈ 𝑋 such that 𝑥 ∼_ℐ𝑦 and 𝑦 ∼_ℐ𝑧. Then we have

𝑥 ∙ 𝑦, 𝑥 ∗ 𝑦, 𝑦 ∙ 𝑧,𝑦 ∗ 𝑧 ∈ ℐand 𝑦 ∙ 𝑥, 𝑦 ∗ 𝑥, 𝑧 ∙ 𝑦,𝑧 ∗ 𝑦 ∈ ℐ. Since ℐ is a pseudo-UP ideal of 𝑋, we have

(𝑦 ∙ 𝑧) ∗ [(𝑥 ∙ 𝑦) ∗ (𝑥 ∙ 𝑧)] = 0 ∈ ℐ and (𝑦 ∗ 𝑧) ∙ [(𝑥 ∗ 𝑦) ∙ (𝑥 ∗ 𝑧)] = 0 ∈ ℐ. c b a 0 ∙ c b a 0 0 c b 0 0 a c 0 0 0 b 0 b 0 0 c c b a 0 ∗ c b a 0 0 c b 0 0 a c 0 a 0 b 0 b a 0 c

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Since, (𝑦 ∙ 𝑧), (𝑦 ∗ 𝑧) ∈ ℐthen by Proposition 2.5, (𝑥 ∙ 𝑦) ∗ (𝑥 ∙ 𝑧), (𝑥 ∗ 𝑦) ∙ (𝑥 ∗ 𝑧) ∈ ℐ. Again, since (𝑥 ∙ 𝑦), (𝑥 ∗ 𝑦) ∈ ℐthen by Proposition 2.5, (𝑥 ∙ 𝑧), (𝑥 ∗ 𝑧) ∈ ℐ. Similarly, ℐ is a pseudo-UP ideal of 𝑋,

(𝑦 ∙ 𝑥) ∗ [(𝑧 ∙ 𝑦) ∗ (𝑧 ∙ 𝑥)] = 0 ∈ ℐ and (𝑦 ∗ 𝑥) ∙ [(𝑧 ∗ 𝑦) ∙ (𝑧 ∗ 𝑥)] = 0 ∈ ℐ.

Since (𝑦 ∙ 𝑥), (𝑦 ∗ 𝑥) ∈ ℐ then by Proposition 2.5, (𝑧 ∙ 𝑦) ∗ (𝑧 ∙ 𝑥), (𝑧 ∗ 𝑦) ∙ (𝑧 ∗ 𝑥) ∈ ℐ. Again, since (𝑧 ∙ 𝑦), (𝑧 ∗ 𝑦) ∈ ℐthen by Proposition 2.5, (𝑧 ∙ 𝑥), (𝑧 ∗ 𝑥) ∈ ℐ. Therefore, 𝑥 ∼ℐ𝑦.

Thus, ∼_ℐ is an equivalent relation on 𝑋.

Now, let 𝑥,𝑦, 𝑢,𝑣 ∈ 𝑋 such that 𝑥 ∼_ℐ𝑢 and 𝑦 ∼_ℐ𝑣. Then we have

𝑥 ∙ 𝑢,𝑥 ∗ 𝑢, 𝑦 ∙ 𝑣, 𝑦 ∗ 𝑣 ∈ ℐ and 𝑢 ∙ 𝑥, 𝑢 ∗ 𝑥, 𝑦 ∙ 𝑣, 𝑦 ∗ 𝑣 ∈ ℐ. Since ℐis a pseudo-UP ideal of 𝑋, we have

(𝑣 ∙ 𝑦) ∗ [(𝑥 ∙ 𝑣) ∗ (𝑥 ∙ 𝑦)] = 0 ∈ ℐ and (𝑣 ∗ 𝑦) ∙ [(𝑥 ∗ 𝑣) ∙ (𝑥 ∗ 𝑦)] = 0 ∈ ℐ.

Since, (𝑣 ∙ 𝑦),(𝑣 ∗ 𝑦) ∈ ℐ then by Proposition 2.5, (𝑥 ∙ 𝑣) ∗ (𝑥 ∙ 𝑦), (𝑥 ∗ 𝑣) ∙ (𝑥 ∗ 𝑦) ∈ ℐ. Similarly, since ℐis a pseudo-UP ideal of 𝑋, we have

(𝑦 ∙ 𝑣) ∗ [(𝑥 ∙ 𝑦) ∗ (𝑥 ∙ 𝑣)] = 0 ∈ ℐ and (𝑦 ∗ 𝑣) ∙ [(𝑥 ∗ 𝑦) ∙ (𝑥 ∗ 𝑣)] = 0 ∈ ℐ.

Since (𝑦 ∙ 𝑣),(𝑦 ∗ 𝑣) ∈ ℐ then by Proposition 2.5, (𝑥 ∙ 𝑦) ∗ (𝑥 ∙ 𝑣),(𝑥 ∗ 𝑦) ∙ (𝑥 ∗ 𝑣) ∈ ℐ. Therefore, 𝑥 ∙ 𝑦 ∼_ℐ 𝑢 ∙ 𝑣 and 𝑥 ∗ 𝑦 ∼_ℐ 𝑢 ∗ 𝑣. On the other hand, since ℐ is a pseudo-UP ideal of 𝑋, we have

(𝑢 ∙ 𝑣) ∗ [(𝑥 ∙ 𝑢) ∗ (𝑥 ∙ 𝑣)] = 0 ∈ ℐand (𝑢 ∗ 𝑣) ∙ [(𝑥 ∗ 𝑢) ∙ (𝑥 ∗ 𝑣)] = 0 ∈ ℐ.

Since (𝑥 ∙ 𝑢),(𝑥 ∗ 𝑢) ∈ ℐ, then (𝑢 ∙ 𝑣) ∗ (𝑥 ∙ 𝑣),(𝑢 ∗ 𝑣) ∙ (𝑥 ∗ 𝑣) ∈ ℐ. Similarly, since ℐis a pseudo-UP ideal of 𝑋, we have

(𝑥 ∙ 𝑣) ∗ [(𝑢 ∙ 𝑥) ∗ (𝑢 ∙ 𝑣)] = 0 ∈ ℐ and (𝑥 ∗ 𝑣) ∙ [(𝑢 ∗ 𝑥) ∙ (𝑢 ∗ 𝑣)] = 0 ∈ ℐ.

Since, (𝑢 ∙ 𝑥),(𝑢 ∗ 𝑥) ∈ ℐ then (𝑥 ∙ 𝑣) ∗ (𝑢 ∙ 𝑣), (𝑥 ∗ 𝑣) ∙ (𝑢 ∗ 𝑣) ∈ ℐ. Therefore, 𝑥 ∙ 𝑣 ∼_ℐ 𝑢 ∙ 𝑣 and 𝑥 ∗ 𝑣 ∼_ℐ𝑢 ∗ 𝑣. By transitive of ∼ℐ we have 𝑥 ∙ 𝑦 ∼ℐ 𝑢 ∙ 𝑣 and 𝑥 ∗ 𝑦 ∼ℐ𝑢 ∗ 𝑣. Hence, ∼ℐis a congruence relation on 𝑋.

4. Uniform topology on pseudo-UP algebras

Suppose that 𝑋 is a pseudo-UP algebra and 𝑈, 𝑉 ⊆ 𝑋 × 𝑋, consider the following notations: 𝑈[𝑥] = {𝑦 ∈ 𝑋 ∶ (𝑥, 𝑦) ∈ 𝑈 },

𝑈 ∘ 𝑉 = {(𝑥,𝑦) ∈ 𝑋 × 𝑋| 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑧 ∈ 𝑋, (𝑥, 𝑧) ∈ 𝑈 𝑎𝑛𝑑 (𝑧,𝑦) ∈ 𝑉}, 𝑈−1_{= {(𝑥,𝑦) ∈ 𝑋 × 𝑋| (𝑦,𝑥) ∈ 𝑈},}

Δ = {(𝑥, 𝑥) ∈ 𝑋 × 𝑋| 𝑥 ∈ 𝑋}.

Definition 4.1. [3] A uniformity on X is defined as a collection 𝒦 of subsets of 𝑋 × 𝑋that satisfy the following

conditions: for all 𝑈, 𝑉 ∈ 𝒦, (𝑈₁)∆ ⊆ 𝑈,

(𝑈₂)𝑈−1 ⊆ 𝒦,

(𝑈₃)𝑊 ∘ 𝑊 ⊆ 𝑈 for some 𝑊 ∈ 𝒦, (𝑈₄) 𝑈 ∩ 𝑉 ∈ 𝒦, and

(𝑈₅) if 𝑈 ⊆ 𝑊 ⊆ 𝑋 × 𝑋, then 𝑊 ∈ 𝒦.

Then (𝑋,𝒦)is said to be a uniform space (or uniform stracture).

Definition 4.2. Suppose that 𝛬 is an arbitrary family of pseudo-UP ideals of a pseudo-UP algebra 𝑋, 𝑈 ⊆ 𝑋 × 𝑋

and 𝐴 ⊆ 𝑋, then we define the following sets:

1. 𝑈ℐ= {(𝑥, 𝑦) ∈ 𝑋 × 𝑋: 𝑥 ∙ 𝑦, 𝑥 ∗ 𝑦 ∈ ℐ 𝑎𝑛𝑑 𝑦 ∙ 𝑥,𝑦 ∗ 𝑥 ∈ ℐ},

2. 𝑈ℐ[𝑥] = {𝑦 ∈ 𝑋: (𝑥,𝑦) ∈ 𝑈ℐ}, and 𝑈ℐ[𝐴] = ⋃𝑎∈𝐴𝑈ℐ[𝑎],

3. 𝒦∗_{= {𝑈}

ℐ: ℐ ∈ 𝛬},

Proposition4.3. Let 𝛬 be a family of pseudo-UP ideals of a pseudo-UP-algebra 𝑋, then 𝒦∗ satisfies the conditions (𝑈1) − (𝑈4).

Proof. (𝑈₁): Since ℐ is a pseudo-UP ideal of 𝑋, then for all 𝑥 ∈ 𝑋, we have 𝑥 ∼_ℐ𝑥. Hence, ∆ ⊆ 𝑈_ℐ, for any

𝑈ℐ∈ 𝒦∗.

(𝑈₂): Let 𝑈_ℐ∈ 𝒦∗, we have

(𝑥,𝑦) ∈ (𝑈ℐ)−1⇔ (𝑦,𝑥) ∈ 𝑈ℐ⇔ 𝑦 ∼ℐ𝑥 ⇔ 𝑥 ∼ℐ𝑦 ⇔ (𝑥, 𝑦) ∈ 𝑈ℐ.

Hence, 𝑈−1 ⊆ 𝒦∗.

(𝑈₃):Let 𝑈_ℐ∈ 𝒦∗ and (𝑥, 𝑧) ∈ 𝑈_ℐ∘ 𝑈_ℐ, then there exists 𝑦 ∈ 𝑋 such that (𝑥, 𝑦),(𝑦, 𝑧) ∈ 𝑈_ℐ implies that 𝑥 ∼_ℐ𝑦 and 𝑦 ∼_ℐ𝑧. By transitive of ∼_ℐ we have 𝑥 ∼_ℐ𝑧. Therefore, (𝑥, 𝑧) ∈ 𝑈_ℐ and hence 𝑈_ℐ∘ 𝑈_ℐ⊆ 𝑈_ℐ.

(𝑈₄): Let 𝑈_ℐ, 𝑈_𝒩∈ 𝒦∗. We claim that 𝑈_ℐ∩ 𝑈_𝒩∈ 𝒦∗. Let

(𝑥, 𝑦) ∈ 𝑈ℐ∩ 𝑈𝒩⇔ (𝑥, 𝑦) ∈ 𝑈ℐ 𝑎𝑛𝑑 (𝑥,𝑦) ∈ 𝑈𝒩⇔ 𝑥 ∙ 𝑦, 𝑥 ∗ 𝑦 ∈ ℐ ∩ 𝒩 𝑎𝑛𝑑 𝑦 ∙ 𝑥,

𝑦 ∗ 𝑥 ∈ ℐ ∩ 𝒩 ⇔ 𝑥 ∼ℐ∩ 𝒩𝑦 ⇔ (𝑥,𝑦) ∈ 𝑈ℐ∩ 𝒩.

Therefore, 𝑈_ℐ∩ 𝑈_𝒩= 𝑈_{ℐ∩ 𝒩}. Since, 𝑈_ℐ, 𝑈_𝒩 ∈ 𝛬, then we have 𝑈_ℐ∩ 𝑈_𝒩∈ 𝒦∗ and so 𝑈_{ℐ∩ 𝒩}∈ 𝒦∗. The following example explain that𝒦∗ is not uniform structure.

Example 4.4. Let 𝑋 = {0,𝑎, 𝑏, 𝑐, 𝑑} and the two binary operations ∙ and ∗defined by the following Cayley tables:

Table 3. A pseudo-UP ideal of a pseudo-UPalgebra

Then it is clear that ((𝑋,≤), ∙, ∗, 0) is a pesudo-UP algebra, {0},𝑋, ℐ₁= {0, 𝑎, 𝑏}and ℐ₂= {0, 𝑎, 𝑐} are pseudo-UP ideals of 𝑋. Hence, 𝒦∗= {𝑈_{0}, 𝑈_𝑋, 𝑈_ℐ1, 𝑈_ℐ2} where 𝑈_{0}= 𝛥, 𝑈_𝑋= 𝑋 × 𝑋,

𝑈ℐ1= {(0,0),(𝑎, 𝑎),(𝑏, 𝑏), (𝑐, 𝑐),(𝑑, 𝑑),(0,𝑎),(𝑎, 0),(0,𝑏), (𝑏, 0),(𝑎,𝑏), (𝑏,𝑎)},

and

𝑈ℐ2= {(0,0),(𝑎, 𝑎),(𝑏,𝑏),(𝑐, 𝑐),(𝑑, 𝑑),(0,𝑎),(𝑎, 0),(0,𝑐),(𝑎, 𝑐),(𝑐,𝑎)}.

Let 𝑀 = ℐ₁∩ {𝑑} = {0,𝑎, 𝑏, 𝑑}, then 𝑈_𝑀= 𝑈_ℐ1∪ {(𝑑, 𝑑),(𝑎, 𝑑), (𝑑, 𝑎), (𝑏, 𝑑), (𝑑, 𝑏), (𝑐,𝑑), (𝑑, 𝑐)}. We have 𝑈ℐ1⊆ 𝑈𝑀⊆ 𝑋 × 𝑋. Moreover, 𝑀 ∉ 𝛬 since 𝑑 ∙ 𝑐 = 0 ∈ 𝑀, 𝑑 ∗ 𝑐 = 0 ∈ 𝑀 and 𝑑 ∈ 𝑀 but 𝑐 ∉ 𝑀. Therefore,

𝑈𝑀∉ 𝒦∗.This means that𝒦∗is not satisfying the condition(𝑈5) from Definition 4.1.

Proposition 4.5. Let 𝛬 be a family of pseudo-UP ideals of a pseudo-UP-algebra 𝑋, then(𝑋,𝒦)is a uniform

structure.

Proof. From Proposition 4.3, we obtain that 𝒦satisfying the conditions(𝑈₁) − (𝑈₄). It is enough to show that 𝒦

satisfying(𝑈₅). Let 𝑈 ∈ 𝒦 and 𝑈 ⊆ 𝑉 ⊆ 𝑋 × 𝑋, then there exists a 𝑈_ℐ⊆ 𝑈 ⊆ 𝑉, which means 𝑉 ∈ 𝒦. Hence the proof.

Lemma 4.6. Let 𝑋 be a pseudo-UP algebra and let𝑈,𝑉 ∈ 𝒦 where 𝑈 ⊆ 𝑉, then 𝑈[𝑥] ⊆ 𝑉[𝑥] for every 𝑥 ∈ 𝑋.

Proof. Suppose that 𝑈, 𝑉 ∈ 𝒦 where 𝑈 ⊆ 𝑉 and let𝑦 ∈ 𝑋. Let 𝑏 ∈ 𝑈[𝑦], then (𝑦, 𝑏) ∈ 𝑈 ⊆ 𝑉and so(𝑦,𝑏) ∈ 𝑉.

Thus, 𝑏 ∈ 𝑉[𝑦] and hence 𝑈[𝑦] ⊆ 𝑉[𝑦].

Proposition 4.7. Let (𝑋,𝒦) be a uniform structure, then d c b a 0 ∙ d c b a 0 0 d c b 0 0 a d c 0 0 0 b d 0 b 0 0 c 0 0 0 0 0 d d c b a 0 ∗ d c b a 0 0 d c b 0 0 a d c 0 0 0 b d 0 b 0 0 c 0 0 0 0 0 d

6410

𝒯: = {𝐺 ⊆ 𝑋: ∀ 𝑥 ∈ 𝐺, ∃𝑈 ∈ 𝒦, 𝑈[𝑥] ⊆ 𝐺}, is a topology on 𝑋.

Proof. Let (𝑋,𝒦)be a uniform structure, for all 𝑥 ∈ 𝑋 and 𝑈 ∈ 𝒦, 𝑈[𝑥] ⊆ 𝑋. Hence, 𝑋 ∈ 𝒯 and also 𝜙 ∈ 𝒯 by

definition. Let 𝑥 ∈ ⋃_{𝐺𝑖∈𝒯,𝑖∈𝑀}𝐺_𝑖, then there exists 𝑗 ∈ 𝑀 such that 𝑥 ∈ 𝐺_𝑗. Since 𝐺_𝑗∈ 𝒯, there exist 𝑈_𝑗 ∈ 𝒦 such that 𝑈_𝑗[𝑥] ⊆ 𝐺_𝑗[𝑥]. This implies that 𝑈_𝑗[𝑥] ⊆ ⋃_{𝐺𝑖∈𝒯,𝑖∈𝑀}𝐺_𝑖. Hence, ⋃_{𝐺𝑖∈𝒯,𝑖∈𝑀}𝐺_𝑖∈ 𝒯.

Suppose that 𝐺, 𝐻 ∈ 𝒯 such that 𝑥 ∈ 𝐺 ∩ 𝐻, then there exist 𝑈, 𝑉 ∈ 𝒦 such that 𝑈[𝑥] ∈ 𝐺 and 𝑉[𝑥] ∈ 𝐻. Let 𝑊: = 𝑈 ∩ 𝑉 so by Definition 4.1, 𝑊 ∈ 𝒦. Let 𝑦 ∈ 𝑊[𝑥], then (𝑥, 𝑦) ∈ 𝑈 and (𝑥, 𝑦) ∈ 𝑉. Therefore, 𝑦 ∈ 𝑈[𝑥] and 𝑦 ∈ 𝑉[𝑥]. Hence, 𝑊[𝑥] ⊆ 𝑈[𝑥] ∩ 𝑉[𝑥]. Therefore, we have 𝑊[𝑥] ⊆ 𝑈[𝑥] ⊆ 𝐺 and 𝑊[𝑥] ⊆ 𝑉[𝑥] ⊆ 𝐻. Hence, 𝑊[𝑥] ⊆ 𝐺 ∩ 𝐻 which implies that 𝐺 ∩ 𝐻 ∈ 𝒯. Hence, 𝒯 is a topology on 𝑋.

Note that 𝑈[𝑥] is an open set containing 𝑥 for all 𝑥 ∈ 𝑋. Moreover, we refer the uniform topology obtained by an arbitrary family 𝛬, by 𝒯_Λ and if Λ = ℐ, we refer to it by 𝒯_ℐ.

Definition 4.8.If (𝑋,𝒦) is a uniform structure, the topology 𝒯 is called uniform topology on 𝑋 induced by 𝒦. Example 4.9. From Example 3.3, consider ℐ = {0,𝑎, 𝑐} and Λ = {ℐ}then we have

𝒦∗_{= { 𝑈}

ℐ} = {(𝑥, 𝑦)| 𝑥 ∼ℐ 𝑦}

= {(0,0),(0,𝑎), (𝑎,0),(0,𝑐), (𝑐,0),(𝑎, 𝑐),(𝑐, 𝑎), (𝑎,𝑎), (𝑏, 𝑏), (𝑐,𝑐)}.

Then it is easy to check that (𝑋,𝒦) is a uniform structure, where 𝒦 = {𝑈| 𝑈_ℐ⊆ 𝑈}. Therefore, the open sets are 𝑈ℐ[𝑎] = {0,𝑎, 𝑐}

𝑈ℐ[𝑏] = {𝑏}

𝑈ℐ[𝑐] = {0,𝑎,𝑐}

𝑈ℐ[0] = {0,𝑎, 𝑐}.

From above we obtain that𝒯_ℐ= {𝜙, {𝑏}, {0,𝑎, 𝑐}, 𝑋}. Hence, (𝑋,𝒯_ℐ) is a uniform topological space.

Proposition 4.10.In a pseudo-UP algebra 𝑋, (𝑋,𝒯_Λ)is a TPUP-algebra.

Proof.Suppose that 𝐺, 𝐻 are open sets containing 𝑥 ∙ 𝑦 and 𝑥 ∗ 𝑦 for all 𝑥, 𝑦 ∈ 𝑋. Then there is 𝑈 ∈ 𝒦, such that

𝑈[𝑥 ∙ 𝑦] ⊆ 𝐺, 𝑈[𝑥 ∗ 𝑦] ⊆ 𝐻 and a pseudo-UP ideal ℐ of 𝑋 such that 𝑈ℐ⊆ 𝑈. We claim that the following relation

holds:

𝑈ℐ[𝑥] ∙ 𝑈ℐ[𝑦] ⊆ 𝑈[𝑥 ∙ 𝑦] 𝑎𝑛𝑑 𝑈ℐ[𝑥] ∗ 𝑈ℐ[𝑦] ⊆ 𝑈[𝑥 ∗ 𝑦].

Let 𝑎 ∈ 𝑈_ℐ[𝑥] and 𝑏 ∈ 𝑈_ℐ[𝑦], then we have 𝑥 ∼_ℐ𝑎 and 𝑦 ∼_ℐ𝑏. Since ∼_ℐ is a congruence relation, it follows that 𝑥 ∙ 𝑦 ∼ℐ𝑎 ∙ 𝑏 and 𝑥 ∗ 𝑦 ∼ℐ𝑎 ∗ 𝑏. Thus, (𝑥 ∙ 𝑦, 𝑎 ∙ 𝑏) ∈ 𝑈ℐ ⊆ 𝑈 and (𝑥 ∗ 𝑦, 𝑎 ∗ 𝑏) ∈ 𝑈ℐ⊆ 𝑈. Hence, 𝑎 ∙ 𝑏 ∈ 𝑈ℐ[𝑥 ∙

𝑦] ⊆ 𝑈[𝑥 ∙ 𝑦] and 𝑎 ∗ 𝑏 ∈ 𝑈ℐ[𝑥 ∗ 𝑦] ⊆ 𝑈[𝑥 ∗ 𝑦]. Therefore, 𝑎 ∙ 𝑏 ∈ 𝐺 and 𝑎 ∗ 𝑏 ∈ 𝐻. Clearly, 𝑈ℐ[𝑥] and 𝑈ℐ[𝑦] are

open sets containing 𝑥 and 𝑦 respectively. Hence, (𝑋, 𝒯_Λ) is a TPUP-algebra.

Proposition 4.11. [3] Let 𝑋 be any set and 𝔖 ⊆ 𝒫(𝑋 × 𝑋) be a family where the following conditions hold: for all

𝑈 ∈ 𝔖, 1. Δ ⊆ 𝑈,

2. 𝑈−1_{includes an element of 𝔖, and}

3. there is a𝑉 ∈ 𝒮 such that 𝑉 ∘ 𝑉 ⊆ 𝑈.

So, there is a unique uniformity 𝒰, of which 𝒮 is a subbase.

Proposition 4.12. Let 𝔇: = {𝑈_ℐ: ℐ be a pseudo-UP ideal of a pseudo-UP algebra 𝑋}, then 𝔇 is the subbase for the

uniformity of 𝑋. We refer to its correlating topology by𝔖.

Proof.Clearly 𝔇 satisfies all conditions of Proposition 4.11 because ∼_ℐ is an equivalence relation.

Example 4.13. In Example 4.9, it is clear that (𝑋,𝒯_ℐ) is a TPUP-algebra.

Proposition 4.14.Let ℐ be a pseudo-UP ideal of a pseudo-UP algebra 𝑋. ℐ = {0} if and only if 𝑈_ℐ= 𝑈_{0}.

Proof.Suppose that ℐ ≠ {0}, then there exist 𝑧 ∈ ℐ such that 𝑧 ≠ 0. Since 𝑧 ∙ 0 = 0 ∈ ℐ, 0 ∙ 𝑧 = 𝑧 ∈ ℐ and

𝑧 ∗ 0 = 0 ∈ ℐ,0 ∗ 𝑧 = 𝑧 ∈ ℐ. Hence,0 ∈ 𝑈ℐ[𝑧] and so (𝑧,0) ∈ 𝑈ℐ. On the other hand since 𝑧 ≠ 0, (𝑧,0) ∉ 𝑈{0}.

Proposition 4.15. Let Λ be a family of pseudo-UP ideals of a pseudo-UP algebra 𝑋. Then any pseudo-UP ideal in

the collection Λ is a clopen subset of 𝑋.

Proof.Let ℐ be a pseudo-UP ideal of 𝑋 in Λ and 𝑦 ∈ ℐ𝑐. Then 𝑦 ∈ 𝑈_ℐ[𝑦] and we have ℐ𝑐⊆ ⋃{𝑈_ℐ[𝑦] | 𝑦 ∈ ℐ𝑐}. We claim that 𝑈_ℐ[𝑦] ⊆ ℐ𝑐for all 𝑦 ∈ ℐ𝑐. Let 𝑧 ∈ 𝑈_ℐ[𝑦], then 𝑧 ∼_ℐ𝑦 and so 𝑧 ∙ 𝑦, 𝑧 ∗ 𝑦 ∈ ℐ. If 𝑧 ∈ ℐ then 𝑦 ∈ ℐ, which is a contradiction. Thus, 𝑧 ∈ ℐ𝑐 and we have ⋃{𝑈_ℐ[𝑦]| 𝑦 ∈ ℐ𝑐} ⊆ ℐ𝑐.Hence, ℐ𝑐= ⋃{𝑈_ℐ[𝑦] | 𝑦 ∈ ℐ𝑐}.Since,𝑈_ℐ[𝑦] is an open for any 𝑦 ∈ 𝑋, then ℐ is a closed subset of 𝑋. Next, we have to prove that ℐ = ⋃{𝑈_ℐ[𝑦] | 𝑦 ∈ ℐ}. If 𝑦 ∈ ℐ, then 𝑦 ∈ 𝑈_ℐ[𝑦] and hence ℐ ⊆ ⋃{𝑈_ℐ[𝑦] | 𝑦 ∈ ℐ}.Let 𝑦 ∈ ℐ, if 𝑧 ∈ 𝑈_ℐ[𝑦] then 𝑦 ∼_ℐ𝑧 and so 𝑦 ∙ 𝑧, 𝑦 ∗ 𝑧 ∈ ℐ. Since 𝑦 ∈ ℐ, and ℐ is a pseudo-UP ideal then 𝑧 ∈ ℐ. Hence, we have ⋃{𝑈ℐ[𝑦] | 𝑦 ∈ ℐ} ⊆ ℐ. Hence, ℐ is an open subset of

𝑋.

Proposition 4.16. Let Λ be a family of pseudo-UP ideals of a pseudo-UP algebra 𝑋, then 𝑈_ℐ[𝑥] is clopen subset of

𝑋 for all 𝑥 ∈ 𝑋 and ℐ ∈ Λ.

Proof.We have to prove (𝑈_ℐ[𝑥])𝑐 is open. If 𝑦 ∈ (𝑈_ℐ[𝑥])𝑐, then 𝑥 ∙ 𝑦,𝑥 ∗ 𝑦 ∈ ℐ𝑐 or 𝑦 ∙ 𝑥, 𝑦 ∗ 𝑥 ∈ ℐ𝑐. Let 𝑦 ∙ 𝑥,𝑦 ∗ 𝑥 ∈ ℐ𝑐_{, then by Proposition 4.10 and the proof of Proposition 4.15, we get (𝑈}

ℐ[𝑦] ∙ 𝑈ℐ[𝑥]) ⊆ 𝑈ℐ[𝑦 ∙ 𝑥] ⊆ ℐ𝑐 and

(𝑈ℐ[𝑦] ∗ 𝑈ℐ[𝑥]) ⊆ 𝑈ℐ[𝑦 ∗ 𝑥] ⊆ ℐ𝑐. We claim that 𝑈ℐ[𝑦] ⊆ (𝑈ℐ[𝑥])𝑐. If 𝑧 ∈ 𝑈ℐ[𝑦], then 𝑧 ∙ 𝑥 ∈ (𝑈ℐ[𝑧] ∙ 𝑈ℐ[𝑥]) and

𝑧 ∗ 𝑥 ∈ (𝑈ℐ[𝑧] ∗ 𝑈ℐ[𝑥]). Hence, 𝑧 ∙ 𝑥,𝑧 ∗ 𝑥 ∈ ℐ𝑐 then we have 𝑧 ∈ (𝑈ℐ[𝑥])𝑐, hence (𝑈ℐ[𝑥])𝑐 is open. Thus, 𝑈ℐ[𝑥]

is closed. It is clear that 𝑈_ℐ[𝑥] is open. Therefore, 𝑈_ℐ[𝑥] is clopen subset of 𝑋.

A topological space (𝑋,𝜏) is connected if and only if 𝑋 and 𝜙 are only clopen sets in 𝜏. Thus, we get the following result.

Corollary 4.17.(𝑋,𝒯_Λ) is disconnected space. Proposition 4.18. 𝒯_Λ= 𝒯_𝒩,where 𝒩 = ⋂{ ℐ: ℐ ∈ Λ}.

Proof.Let 𝒦 and 𝒦∗ defined as in Definition 4.1 and 4.2. Now, consider Λ₀= {𝒩} and define 𝒦0∗= {𝒩} 𝑎𝑛𝑑 𝒦0= {𝑈: 𝑈𝒩⊆ 𝑈}.

Let 𝐺 ∈ 𝒯_Λ, so for every𝑥 ∈ 𝐺 there is𝑈 ∈ 𝒦 such that 𝑈[𝑥] ⊆ 𝐺. Since𝒩 ⊆ ℐ, then we have 𝑈_𝒩⊆ 𝑈 _ℐ, for ever y pseudo-UP ideal ℐof Λ. Since, 𝑈 ∈ 𝒦 there isℐ ∈ Λ such that 𝑈 _ℐ⊆ 𝑈. Thus, 𝑈_𝒩[𝑥] ⊆ 𝑈 _ℐ[𝑥] ⊆ 𝐺. Since 𝑈_𝒩∈ 𝒦0, 𝐺 ∈ 𝒯𝒩. Hence, 𝒯Λ⊆ 𝒯𝒩.

Conversely, let 𝐻 ∈ 𝒯_𝒩 then for every𝑥 ∈ 𝐻, there is 𝑈 ∈ 𝒦₀ such that 𝑈[𝑥] ⊆ 𝐻. Thus, 𝑈_𝒩[𝑥] ⊆ 𝐻 and since Λ is closed under intersection 𝒩 ∈ Λ. Then we obtain that 𝑈_𝒩 ∈ 𝒦 and so 𝐻 ∈ 𝒯_Λ. Therefore, 𝒯_𝒩 ⊆ 𝒯_Λ.

Remark 4.19. Let Λ be a family of pseudo-UP ideals of a pseudo-UP algebra 𝑋 and 𝒩 = ⋂{ ℐ: ℐ ∈ Λ}.Then the

following statements hold:

1. By Proposition 4.18, we have𝒯Λ= 𝒯𝒩. For all𝑈 ∈ 𝒦, and for all𝑥 ∈ 𝑋 we get𝑈𝒩[𝑥] ⊆ 𝑈[𝑥]. Hence, 𝒯Λ is

equivalent to {𝐺 ⊆ 𝑋: ∀ 𝑥 ∈ 𝐺,∃ 𝑈_𝒩[𝑥] ⊆ 𝐺}. Therefore, 𝐺 ∈ 𝑋 is an open set if and only if for all 𝑥 ∈ 𝐺, 𝑈𝒩[𝑥] ⊆ 𝐺 if and only if 𝐺 = ⋃𝑥∈𝐺𝑈𝒩[𝑥].

2. By (1) we get𝑈𝒩[𝑥]is a minimal open set containing 𝑥 for all 𝑥 ∈ 𝑋.

3. Let 𝔅𝒩= {𝑈𝒩[𝑥]: 𝑥 ∈ 𝑋}. By (1), and (2) it is easy to show that 𝔅𝒩 is a base of 𝒯𝒩.

Proposition 4.20. Let ℐ and 𝒩 be two pseudo-UP ideals of a pseudo-UP algebra 𝑋. Then 𝒯_ℐ⊆ 𝒯_𝒩 if and only if 𝒩 ⊆ ℐ.

Proof.Let 𝒩 ⊆ ℐ, and consider:

Λ1= {ℐ}, 𝒦1∗= {𝑈 ℐ}, 𝒦1= {𝑈: 𝑈ℐ⊆ 𝑈} andΛ2= {𝒩},𝒦2∗= {𝑈 𝒩}, 𝒦2= {𝑈: 𝑈𝒩⊆ 𝑈}.

Let 𝐺 ∈ 𝒯_ℐ, then for all 𝑥 ∈ 𝐺, there exist 𝑈 ∈ 𝒦₁ such that 𝑈[𝑥] ∈ 𝐺. Since 𝒩 ⊆ ℐ, then 𝑈_𝒩 ⊆ 𝑈 _ℐ. Since 𝑈 ℐ[𝑥] ⊆ 𝐺, we have 𝑈𝒩[𝑥] ⊆ 𝐺. Then, 𝑈𝒩 ∈ 𝒦2 and thus 𝐺 ∈ 𝒯𝒩.

Conversely, let 𝒯_ℐ⊆ 𝒯_𝒩. Suppose that 𝑎 ∈ 𝒩\ ℐ, since ℐ ∈ 𝒯_ℐ by assumption we have ℐ ∈ 𝒯_𝒩. Then for all 𝑥 ∈ ℐ, there exist 𝑈 ∈ 𝒦₂ such that 𝑈[𝑥] ⊆ ℐ, and so 𝑈_𝒩[𝑥] ⊆ ℐ. Then, 𝑈_𝒩[0] ⊆ ℐwe have 𝑎 ∙ 0 = 0 ∈ 𝒩, 0 ∙ 𝑎 = 𝑎 ∈ 𝒩, 𝑎 ∗ 0 = 0 ∈ 𝒩 and 0 ∗ 𝑎 = 𝑎 ∈ 𝒩. Thus, 𝑎 ∼𝒩0, and so 𝑎 ∈ 𝑈𝒩[0]. Therefore 𝑎 ∈ ℐ which is a

contradiction.

A uniform structure (𝑋,𝒦)is called totally bounded if for every 𝑈 ∈ 𝒦, there exists 𝑥₁,𝑥₂,.. . , 𝑥_𝑛∈ 𝑋 such that 𝑋 = ⋃ 𝑈[𝑥𝑛𝑖=1 𝑖]. Moreover, (𝑋,𝒦)is compact if for every open cover of 𝑋 has a finite subcover.

6412 Proposition 4.21. Let ℐ be a pseudo-UP ideal of a pseudo-UP algebra 𝑋. Then the following statements are

equivalent:

1. (𝑋,𝒯ℐ) is compact.

2. (𝑋,𝒯ℐ)is totally bounded.

3. There exist 𝑃 = {𝑥1,𝑥2,. . ., 𝑥𝑛} ⊆ 𝑋 such that for every 𝑎 ∈ 𝑋 there exist 𝑥𝑖∈ 𝑃 with 𝑎 ∙ 𝑥𝑖,𝑎 ∗ 𝑥𝑖∈ ℐ,

and 𝑥_𝑖∙ 𝑎, 𝑥_𝑖∗ 𝑎 ∈ ℐ.

Proof.(1 ⇒ 2): Obvious.

(2 ⇒ 3):Let𝑈ℐ∈ 𝒦. Since (𝑋,𝒯ℐ)is totally bounded, so there exists 𝑥1,𝑥2,. . ., 𝑥𝑛∈ ℐ such that 𝑋 = ⋃ 𝑈[𝑥𝑛𝑖=1 𝑖].

Now, let 𝑎 ∈ 𝑋, so there exist 𝑥_𝑖 such that 𝑎 ∈ ⋃ 𝑈[𝑥𝑛_{𝑖=1 𝑖}]. Therefore, 𝑎 ∙ 𝑥𝑖,𝑎 ∗ 𝑥𝑖∈ ℐ 𝑎𝑛𝑑 𝑥𝑖∙ 𝑎, 𝑥𝑖∗ 𝑎 ∈ ℐ.

(3 ⇒ 1): By assumption there exist 𝑥𝑖∈ 𝑃 with 𝑎 ∙ 𝑥𝑖,𝑎 ∗ 𝑥𝑖∈ ℐ, and 𝑥𝑖∙ 𝑎, 𝑥𝑖∗ 𝑎 ∈ ℐ for all 𝑎 ∈ 𝑋.Hence, we get

𝑎 ∈ 𝑈ℐ[𝑥𝑖] and therefore 𝑋 = ⋃ 𝑈𝑛𝑖=1 ℐ[𝑥𝑖]. Now, let 𝑋 = ⋃𝛼∈𝑀𝑂𝛼 where 𝑂𝛼is an open set in 𝑋 for each 𝛼 ∈ 𝑀,

then for every 𝑥_𝑖∈ 𝑋 there exists 𝑥_𝑖∈ 𝑂_𝛼𝑖. Since, 𝑂_𝛼𝑖is open then 𝑈_ℐ[𝑥_𝑖] ⊆ 𝑂_𝛼𝑖 and so 𝑋 = ⋃ 𝑈𝑛_{𝑖=1 ℐ}[𝑥_𝑖]⊆ ⋃ 𝑂𝑛𝑖=1 𝛼𝑖. Therefore, 𝑋 = ⋃ 𝑂𝑛𝑖=1 𝛼𝑖and hence(𝑋,𝒯ℐ) is compact.

Proposition 4.22. Let ℐ be a pseudo-UP ideal of a pseudo-UP algebra 𝑋 such that ℐ𝑐 is finite (𝑋,𝒯_ℐ) is compact.

Proof.Suppose that 𝑋 = ⋃_𝛼∈𝑀𝑂_𝛼 where 𝑂_𝛼is an open set in 𝑋 for each𝛼 ∈ 𝑀, and let ℐ𝑐= {𝑥₁,𝑥₂,. . . ,𝑥_𝑛}. Then there exists 𝛼,𝛼₁,𝛼₂,. .. , 𝛼_𝑛 such that 0 ∈ 𝑂_𝛼, 𝑥₁∈ 𝑂_𝛼1,𝑥₂∈ 𝑂_𝛼2,. . . ,𝑥_𝑛∈ 𝑂_𝛼𝑛. Then 𝑈_ℐ[0] ⊆ 𝑂_𝛼, but 𝑈_ℐ[0] = ℐ. Hence, 𝑋 = ⋃ 𝑂𝑛_{𝑖=1 𝛼𝑖}∪ 𝑂_𝛼.

Proposition 4.23.Let ℐ be a pseudo-UP ideal of a pseudo-UP algebra 𝑋, thenℐ is compact in(𝑋,𝒯_ℐ).

Proof.Suppose that 𝑈_ℐ[𝑥] ⊆ ⋃_𝛼∈𝑀𝑂_𝛼 where 𝑂_𝛼 is an open set in 𝑋 for each 𝛼 ∈ 𝑀. Since, 0 ∈ ℐ then there exist 𝛼 ∈ 𝑀 such that 0 ∈ 𝑂𝛼. Then ℐ = 𝑈ℐ[0] ⊆ 𝑂𝛼 and hence ℐ is a compact set in (𝑋, 𝒯ℐ).

Proposition 4.24.Let ℐ be a pseudo-UP ideal of a pseudo-UP algebra 𝑋, then for all 𝑥 ∈ 𝑋,𝑈_ℐ[𝑥]is compact

in(𝑋, 𝒯_ℐ).

Proof.Suppose that for all 𝑥 ∈ 𝑋, 𝑈_ℐ[𝑥] ⊆ ⋃_𝛼∈𝑀𝑂_𝛼where 𝑂_𝛼 is an open set in 𝑋 for each 𝛼 ∈ 𝑀.Since,𝑥 ∈ 𝑈_ℐ[𝑥], then there exist 𝛼 ∈ 𝑀such that 𝑥 ∈ 𝑂_𝛼. Thus, 𝑈_ℐ[𝑥] ⊆ 𝑂_𝛼. Therefore, 𝑈_ℐ[𝑥] is compact set in (𝑋, 𝒯_ℐ).

Proposition 4.25.Let ℐ be a pseudo-UP ideal of a pseudo-UP algebra 𝑋. Then (𝑋,𝒯_Λ) is a discrete topology if and

only if 𝑈_ℐ[𝑥] = {𝑥} for all 𝑥 ∈ 𝑋.

Proof.Suppose that𝒯_Λ is a discrete topology on 𝑋.If for any ℐ ∈ 𝛬, then there exists 𝑥 ∈ 𝑋 such that 𝑈_ℐ[𝑥] ≠ {𝑥}. Let 𝒩 = ∩ {ℐ ∶ ℐ ∈ 𝛬}, then 𝒩 ∈ 𝛬 so there exist 𝑥₀∈ 𝑋 such that 𝑈_𝒩[𝑥₀] ≠ {𝑥₀}. It follows there is a 𝑦₀∈ 𝑈ℐ[𝑥0] and 𝑥0≠ 𝑦0. By Remark 4.19, we have 𝑈𝒩[𝑥0] is a minimal open set containing 𝑥0. Hence, {𝑥0} is not

open subset of 𝑋 which is a contradiction.

Conversely, for all 𝑥 ∈ 𝑋, there exists ℐ ∈ 𝛬 such that 𝑈_ℐ[𝑥] = {𝑥}. Hence, {𝑥} is open subset of 𝑋. Thus, (𝑋,𝒯_Λ) is a discrete topology.

Proposition 4.26.Let(𝑋,𝒯_Λ)be the topological space where 𝛬 is a family of pseudo-UP ideals of a pseudo-UP

algebra 𝑋 and ℐ ∈ 𝛬. Then for any 𝐴 ⊆ 𝑋, 𝑐𝑙(𝐴) = ∩ {𝑈_ℐ[𝐴]: 𝑈_ℐ∈ 𝒦∗}.

Proof.Let 𝑥 ∈ 𝑐𝑙(𝐴), we have 𝑈_ℐ[𝑥] is an open set containing 𝑥 and so 𝑈_ℐ[𝑥] ∩ 𝐴 ≠ 𝜙, for everyℐ ∈ 𝛬. Thus,

there exist 𝑦 ∈ 𝐴 such that 𝑦 ∈ 𝑈_ℐ[𝑥] and so (𝑥, 𝑦) ∈ 𝑈_ℐ for everyℐ ∈ 𝛬. Therefore, 𝑥 ∈ 𝑈_ℐ[𝑦] ⊆ 𝑈_ℐ[𝐴] for everyℐ ∈ 𝛬.

Conversely, let 𝑥 ∈ 𝑈_ℐ[𝐴] for everyℐ ∈ 𝛬, so there exist 𝑦 ∈ 𝐴 such that 𝑥 ∈ 𝑈_ℐ[𝑦] and hence 𝑈_ℐ[𝑦] ∩ 𝐴 ≠ 𝜙. Thus, 𝑥 ∈ 𝑐𝑙(𝐴).

Proposition 4.27.Let 𝛬 be a family of pseudo-UP ideals of a pseudo-UP algebra 𝑋, and 𝑊 be an open set

containing 𝐾 where 𝐾is a compact subset of 𝑋. Then 𝐾 ⊆ 𝑈_ℐ[𝐾] ⊆ 𝑊.

Proof.Let 𝑊 be an open set containing 𝐾, then for all 𝑘 ∈ 𝐾 we get𝑈_ℐ𝑘[𝑘] ⊆ 𝑊 for some 𝑈_ℐ𝑘∈ 𝛬. Hence,

𝐾 ⊆ ⋃ 𝑈ℐ𝑘 ℐ𝑘[𝑘]⊆ 𝑊. Since 𝐾 is a compact then there exists 𝑘1,𝑘2,.. . , 𝑘𝑛 such that 𝐾 ⊆ 𝑈ℐ𝑘1[𝑘1] ∪ 𝑈ℐ𝑘2[𝑘2] ∪

.. .∪ 𝑈ℐ_𝑘𝑛[𝑘𝑛]. Take ℐ = ⋂ ℐ𝑛𝑖=1 𝑘𝑖. We claim that 𝑈ℐ𝑘[𝑘] ⊆ 𝑊 for all 𝑘 ∈ 𝐾. Let 𝑘 ∈ 𝐾, so there exists 1 ≤ 𝑖 ≤ 𝑛

such that 𝑘 ∈ 𝑈_ℐ𝑘𝑖[𝑘_𝑖] and so 𝑘 ∼_ℐ𝑘𝑖𝑘_𝑖. Let 𝑦 ∈ 𝑈_ℐ[𝑘] then 𝑦 ∼_ℐ𝑘 and so 𝑦 ∼_ℐ𝑘𝑖 𝑘_𝑖. Hence, 𝑦 ∈ 𝑈_ℐ𝑘𝑖[𝑘_𝑖] ⊆ 𝑊 and so 𝑈_ℐ[𝐾] ⊆ 𝑊 for all 𝑘 ∈ 𝐾. Therefore, 𝐾 ⊆ 𝑈_ℐ[𝐾] ⊆ 𝑊.

Proposition 4.28.Let 𝛬 be a family of pseudo-UP ideals of a pseudo-UP algebra 𝑋, and let 𝐾,𝑃 ⊆ 𝑋such that 𝐾 is

a compact and 𝑃is a closed. If 𝐾 ∩ 𝑃 = 𝜙, then 𝑈_ℐ[𝐾] ∩ 𝑈_ℐ[𝑃] = 𝜙 for all ℐ ∈ 𝛬.

Proof.Let𝐾 ∩ 𝑃 = 𝜙 and 𝑃 be a closed, 𝑋\𝑃 is an open set containing 𝐾. By Proposition 4.27, there exists ℐ ∈ 𝛬

such that 𝑈_ℐ[𝐾] ⊆ 𝑋\𝑃. Suppose that 𝑈_ℐ[𝐾] ∩ 𝑈_ℐ[𝑃] ≠ 𝜙, then there exist 𝑦 ∈ 𝑋 such that 𝑦 ∈ 𝑈_ℐ[𝑘] and 𝑦 ∈ 𝑈ℐ[𝑝] for all 𝑘 ∈ 𝐾 and 𝑝 ∈ 𝑃. Hence, 𝑘 ∼ℐ𝑝 and so 𝑝 ∈ 𝑈ℐ[𝑘] ⊆ 𝑈ℐ[𝐾] which is a contradiction with the fact

𝑈ℐ[𝐾] ⊆ 𝑋\𝑃. Hence, 𝑈ℐ[𝐾] ∩ 𝑈ℐ[𝑃] = 𝜙.

From Proposition 2.5, we obtain that every pseudo-UP ideal is a pseudo-UP filter in a pseudo-UP algebra 𝑋, then we have the following result:

Corollary 4.29. Let ((𝑋, ≤), ∙, ∗, 0, 𝜏)be a TPUP-algebra and ℐ₀ is a minimal open set containing 0, then ℐ₀ is a pseudo-UP ideal of 𝑋.

Proposition 4.30.Let ((𝑋, ≤), ∙, ∗, 0, 𝜏)be a TPUP-algebra and(𝑋, 𝒯_ℐ0) be a uniform topology induced by ℐ₀. Then 𝜏 is finer than 𝒯_ℐ0.

Proof. Suppose that 𝑀_𝑦 is the minimal open set in 𝜏 containing 𝑦. We have to show that 𝑈_ℐ0[𝑥] = ⋃_{𝑦∈𝑈ℐ0[𝑥]}𝑀_𝑦 for each 𝑥 ∈ 𝑋. Let 𝑦 ∈ 𝑈_ℐ0[𝑥] and 𝑧 ∈ 𝑀_𝑦. If 𝑧 ∙ 𝑦, 𝑧 ∗ 𝑦 ∉ ℐ₀or 𝑦 ∙ 𝑧, 𝑦 ∗ 𝑧 ∉ ℐ₀, then by Proposition 2.12, 𝑧 ∉ 𝑀𝑦. Therefore, 𝑧 ∙ 𝑦, 𝑧 ∗ 𝑦 ∈ ℐ0or 𝑦 ∙ 𝑧, 𝑦 ∗ 𝑧 ∈ ℐ0. Since, (𝑦 ∙ 𝑧) ∗ [(𝑥 ∙ 𝑦) ∗ (𝑥 ∙ 𝑧)] = 0 ∈ ℐ0, (𝑦 ∗ 𝑧) ∙ [(𝑥 ∗

𝑦) ∙ (𝑥 ∗ 𝑧)] = 0 ∈ ℐ0 and (𝑥 ∙ 𝑦), (𝑥 ∗ 𝑦) ∈ ℐ0, then we have (𝑦 ∙ 𝑧) ∗ (𝑥 ∙ 𝑧), (𝑦 ∗ 𝑧) ∙ (𝑥 ∗ 𝑧) ∈ ℐ0. Again, since

𝑦 ∙ 𝑧, 𝑦 ∗ 𝑧 ∈ ℐ0 then by Proposition 2.5, 𝑥 ∙ 𝑧, 𝑥 ∗ 𝑧 ∈ ℐ0. By the same way we can get 𝑧 ∙ 𝑥, 𝑧 ∗ 𝑥 ∈ ℐ0. Hence,

𝑧 ∈ 𝑈ℐ0[𝑥]. Thus, 𝑀𝑦⊆ 𝑈ℐ0[𝑥] for every 𝑦 ∈ 𝑈ℐ0[𝑥] and hence ⋃𝑦∈𝑈ℐ0[𝑥]𝑀𝑦⊆ 𝑈ℐ0[𝑥]. The converse is clear. Proposition 4.31.Let ((𝑋, ≤), ∙, ∗, 0, 𝜏)be a TPUP-algebra and(𝑋, 𝒯_ℐ0) be a uniform topology induced by ℐ₀.If there exist 𝑈 ∈ 𝜏 such that 𝑈 ∉ 𝒯_ℐ0, then there exist 𝑥 ∈ 𝑈 and 𝑦 ∈ 𝑈_ℐ0[𝑥] such that 𝑦 ∉ 𝑈 and the following statements hold: for all 𝑎 ∈ ℐ₀,

1. 𝑥,𝑦 ∉ ℐ0.

2. 𝑎 ∙ 𝑦, 𝑎 ∗ 𝑦 ∉ 𝑈ℐ0[𝑥] ∩ 𝑈.

3. If 𝑑 ∈ 𝑈ℐ0[𝑥] ∩ 𝑈, then 𝑎 ∙ 𝑑 ≠ 𝑦 and 𝑎 ∗ 𝑑 ≠ 𝑦.

Proof.

1. Let 𝑥 ∈ ℐ0, then by Proposition 2.11, ℐ0⊆ 𝑈. Since 𝑥 ∈ ℐ0, 𝑦 ∈ 𝑈ℐ0[𝑥] and ℐ0is a pseudo-UP ideal, then

𝑦 ∈ ℐ0⊆ 𝑈 which is a contradiction. Now, let 𝑦 ∈ ℐ0 and since 𝑦 ∈ 𝑈ℐ0[𝑥]and ℐ0 is a pseudo-UP ideal,

then 𝑥 ∈ ℐ₀ which is a contradiction.

2. Suppose there exist 𝑎 ∈ 𝑈ℐ0[𝑥] such that 𝑎 ∙ 𝑦,𝑎 ∗ 𝑦 ∈ 𝑈ℐ0[𝑥] ∩ 𝑈, then there exist two open sets 𝑉 and 𝑊

containing 𝑎 and 𝑦 respectively such that 𝑉 ∙ 𝑊 ⊆ 𝑈_ℐ0[𝑥] ∩ 𝑈 and 𝑉 ∗ 𝑊 ⊆ 𝑈_ℐ0[𝑥] ∩ 𝑈. By Proposition 2.11, ℐ₀⊆ 𝑉 and so 𝑦 = 0 ∙ 𝑦 ∈ ℐ₀∙ 𝑊 ⊆ 𝑈_ℐ0[𝑥] ∩ 𝑈 and 𝑦 = 0 ∗ 𝑦 ∈ ℐ₀∗ 𝑊 ⊆ 𝑈_ℐ0[𝑥] ∩ 𝑈. Therefore, 𝑦 ∈ 𝑈ℐ0[𝑥] ∩ 𝑈 which is a contradiction.

3. Suppose that there exist 𝑎 ∈ 𝑈ℐ0[𝑥]such that 𝑎 ∙ 𝑑 = 𝑦 and 𝑎 ∗ 𝑑 = 𝑦 for some 𝑑 ∈ 𝑈ℐ0[𝑥] ∩ 𝑈. Since

0 ∙ 𝑑 = 𝑑 ∈ 𝑈ℐ0[𝑥] ∩ 𝑈 and 0 ∗ 𝑑 = 𝑑 ∈ 𝑈ℐ0[𝑥] ∩ 𝑈, then there exist two open sets 𝑉 and 𝑊 containing 0

and 𝑑 such that 𝑉 ∙ 𝑊 ⊆ 𝑈_ℐ0[𝑥] ∩ 𝑈 and 𝑉 ∗ 𝑊 ⊆ 𝑈_ℐ0[𝑥] ∩ 𝑈. Then we have 𝑦 = 𝑎 ∙ 𝑑 ∈ ℐ₀∙ 𝑊 ⊆ 𝑈ℐ0[𝑥] ∩ 𝑈 and 𝑦 = 𝑎 ∗ 𝑑 ∈ ℐ0∗ 𝑊 ⊆ 𝑈ℐ0[𝑥] ∩ 𝑈. Therefore, 𝑦 ∈ 𝑈ℐ0[𝑥] ∩ 𝑈which is a contradiction. Proposition 4.32.Let ((𝑋, ≤), ∙, ∗, 0, 𝜏)be a TPUP-algebra and(𝑋, 𝒯_ℐ0) be a uniform topology induced by ℐ₀. If 𝒯ℐ0⊊ 𝜏,then there exist a non-empty 𝑈 ∈ 𝜏 such that 𝑈 ⊊ 𝑈ℐ0[𝑥] for some 𝑥 ∉ 𝑋\ℐ0.

Proof. Suppose that 𝒯_ℐ0⊊ 𝜏, then there exist 𝑉 ∈ 𝜏such that 𝑉 ∉ 𝒯_ℐ0. Since (𝑋,𝒯_ℐ0)is a uniform topology, then there exist 𝑥 ∈ 𝑉 such that 𝑈_ℐ0[𝑥] ⊈ 𝑉. Therefore, 𝑈_ℐ0[𝑥] ∩ 𝑉 ⊈ 𝑈_ℐ0[𝑥]. Take 𝑈 = 𝑈_ℐ0[𝑥] ∩ 𝑉, then 𝑈 ∈ 𝜏 and 𝑈 ⊈ 𝑈ℐ0[𝑥]. If 𝑥 ∈ ℐ0, then 𝑈ℐ0[𝑥] = ℐ0. Hence, 𝑥 ∈ 𝑈 and by Proposition 2.11, 𝑈ℐ0[𝑥] = ℐ0⊆ 𝑈 which is a

contradiction.

Proposition 4.33. [9] Let 𝑓: 𝑋 ⟶ 𝑌 be a pseudo-UP homomorphism between two pseudo-UP algebras 𝑋 and 𝑌.

Then the following statements hold:

1. If ℐ is a pseudo-UP ideal in 𝑌, then𝑓−1_{(ℐ) is a pseudo-UP ideal in 𝑋.}

6414 Proposition 4.34.Let 𝑓: 𝑋 ⟶ 𝑌 be a pseudo-UP isomorphism between two pseudo-UP algebras 𝑋,𝑌, and let ℐ is a

pseudo-UP ideal in 𝑌, then for all 𝑥₁,𝑥₂∈ 𝑋,

(𝑥1,𝑥2) ∈ 𝑈𝑓−1_(ℐ)⇔ (𝑓(𝑥₁),𝑓(𝑥₂)) ∈ 𝑈_ℐ.

Proof. For all 𝑥₁,𝑥₂∈ 𝑋, we have

(𝑥1,𝑥2) ∈ 𝑈𝑓−1_(ℐ)⇔ 𝑥₁∼_𝑓−1_(ℐ)𝑥₂⇔ 𝑓(𝑥₁) ∼_ℐ 𝑓(𝑥₂) ⇔ (𝑓(𝑥₁),𝑓(𝑥₂)) ∈ 𝑈_ℐ.

Proposition 4.35.Let 𝑓: 𝑋 ⟶ 𝑌 be a pseudo-UP isomorphism between two pseudo-UP algebras 𝑋,𝑌, and let ℐ is a

pseudo-UP ideal in 𝑌. Then the following statements hold: for all 𝑥 ∈ 𝑋 and for all 𝑦 ∈ 𝑌, 1. 𝑓(𝑈_𝑓−1_(ℐ)[𝑥]) = 𝑈_ℐ[𝑓(𝑥)].

2. 𝑓−1_(𝑈

ℐ[𝑦]) = 𝑈𝑓−1_(ℐ)[𝑓−1(𝑦)].

Proof.

1. Let 𝑦 ∈ 𝑓(𝑈𝑓−1_(ℐ)[𝑥]), then there exist𝑥₁∈ 𝑈_𝑓−1_(ℐ)[𝑥] such that 𝑦 = 𝑓(𝑥₁). It follows that

𝑥 ∼_𝑓−1_(ℐ)𝑥₁⟹ 𝑓(𝑥) ∼_ℐ 𝑓(𝑥₁) ⟹ 𝑓(𝑥) ∼_ℐ𝑦 ⟹ 𝑦 ∈ 𝑈_ℐ[𝑓(𝑥)]. Conversely, let 𝑦 ∈ 𝑈_ℐ[𝑓(𝑥)] ⟹ 𝑓(𝑥) ∼_ℐ𝑦 ⟹ 𝑓−1(𝑓(𝑥) ∼_ℐ𝑦) ⟹ 𝑥 ∼_𝑓−1_(ℐ)𝑓−1(𝑦) ⟹ 𝑓−1(𝑦) ∈ 𝑈_𝑓−1_(ℐ)[𝑥] ⟹ 𝑦 ∈ 𝑓(𝑈_𝑓−1_(ℐ)[𝑥]). 2. Let𝑥 ∈ 𝑓−1_(𝑈 ℐ[𝑦]) ⇔ 𝑓(𝑥) ∈ 𝑈ℐ[𝑦] ⇔ 𝑓(𝑥) ∼ℐ𝑦 ⇔ 𝑓−1(𝑓(𝑥) ∼ℐ𝑦) ⇔ 𝑥 ∼𝑓−1(ℐ) 𝑓−1(𝑦) ⇔ 𝑥 ∈ 𝑈_𝑓−1_(ℐ)[𝑓−1(𝑦)].

Proposition 4.36.Let 𝑓: 𝑋 ⟶ 𝑌 be a pseudo-UP isomorphism between two pseudo-UP algebras 𝑋,𝑌, and let ℐ is a

pseudo-UP ideal in 𝑌. Then 𝑓 is homeomorphism map from (𝑋, 𝒯_𝑓−1_(ℐ)) to (𝑌,𝒯_ℐ).

Proof.First, we have to show that 𝑓 is continuous. Let 𝐴 ∈ 𝒯_ℐ then by Remark 4.19, we get 𝐴 = ⋃_𝑎∈𝐴𝑈_ℐ[𝑎]. It follows that 𝑓−1_{(𝐴) = 𝑓}−1_{(⋃ 𝑈} ℐ[𝑎] 𝑎∈𝐴 ) = ⋃ 𝑓 −1_(𝑈 ℐ[𝑎]) 𝑎∈𝐴 .

We claim that if 𝑏 ∈ 𝑓−1(𝑈_ℐ[𝑎]), then we have 𝑈_𝑓−1_(ℐ)[𝑏] ⊆ 𝑓−1(𝑈_ℐ[𝑎]). Now, let 𝑐 ∈ 𝑈_𝑓−1_(ℐ)[𝑏], then

𝑐 ∼_𝑓−1_(ℐ)𝑏 and so 𝑓(𝑐) ∼_ℐ 𝑓(𝑏). Since 𝑓(𝑏) ∈ 𝑈_ℐ[𝑎]we have 𝑓(𝑏) ∼_ℐ𝑎. Therefore, 𝑓(𝑐) ∼_ℐ𝑎and hence

𝑓(𝑐) ∈ 𝑈ℐ[𝑎]. Thus, 𝑐 ∈ 𝑓−1(𝑈ℐ[𝑎]) and so

𝑓−1_(𝑈

ℐ[𝑎]) = ⋃ 𝑈𝑓−1_(ℐ)[𝑏]

𝑏∈𝑓−1_{(𝑈ℐ[𝑎])} ∈ 𝒯𝑓−1(ℐ).

Therefore, 𝑓−1(𝐴) = 𝑓−1(⋃_𝑎∈𝐴𝑈_ℐ[𝑎]) = ⋃_𝑎∈𝐴𝑓−1(𝑈_ℐ[𝑎]) ∈ 𝒯_𝑓−1_(ℐ)and hence 𝑓 is continuous.

Finally, we have to show that 𝑓 is an open map. Let 𝐴 be an open in (𝑋,𝒯_𝑓−1_(ℐ)). We claim that 𝑓(𝐴) is an open

set in (𝑌,𝒯_ℐ). Let 𝑎 ∈ 𝑓(𝐴) we will have to show that 𝑈_ℐ[𝑎] ⊆ 𝑓(𝐴). Now, for all 𝑏 ∈ 𝑈_ℐ[𝑎] we have 𝑏 ∼_ℐ𝑎. By Proposition 4.34, we have 𝑓−1(𝑏) ∼_𝑓−1_(ℐ)𝑓−1(𝑎). Hence, 𝑓−1(𝑏) ∈ 𝑈_𝑓−1_(ℐ)[𝑓−1(𝑎)]. Since 𝑓 is a one-to-one and

𝑎 ∈ 𝑓(𝐴) then we have 𝑓−1_{(𝑎) ∈ 𝐴. By Remark 4.19, we get that 𝑈}

𝑓−1(ℐ)[𝑓−1(𝑎)] ⊆ 𝐴 and hence 𝑓−1(𝑏) ∈

𝐴implies that 𝑏 ∈ 𝑓(𝐴). Therefore, 𝑈ℐ[𝑎] ⊆ 𝑓(𝐴)and thus 𝑓 is an open map.

5. New topology and related result

In this section we will give a filter base on 𝑋 to generate a topology on 𝑋 where 𝑋 is a pseudo-UP algebra. Let 𝑉 ⊆ 𝑋 and 𝑎 ∈ 𝑋 we define 𝑉(𝑎) as following:

𝑉(𝑎) = {𝑥 ∈ 𝑋: 𝑥 ∙ 𝑎, 𝑥 ∗ 𝑎 ∈ 𝑉 𝑎𝑛𝑑 𝑎 ∙ 𝑥,𝑎 ∗ 𝑥 ∈ 𝑉}. Obviously𝑉(𝑎) ⊆ 𝑈(𝑎) when 𝑉 ⊆ 𝑈 ⊆ 𝑋.

Proposition 5.1.Let 𝑋 be a pseudo-UP algebra satisfying𝑥 ∙ (𝑦 ∙ 𝑧) = 𝑦 ∙ (𝑥 ∙ 𝑧), and 𝑥 ∗ (𝑦 ∗ 𝑧) = 𝑦 ∗ (𝑥 ∗ 𝑧) for

all 𝑥,𝑦,𝑧 ∈ 𝑋 and 𝛺be a filter base satisfying the following conditions: 1. For every 𝑣 ∈ 𝑉 ∈ 𝛺, there exist 𝑈 ∈ 𝛺 such that 𝑈(𝑣) ⊆ 𝑉.

2. If 𝑝,𝑞 ∈ 𝑉 ∈ 𝛺 and 𝑝 ∙ (𝑞 ∙ 𝑥) = 0,𝑝 ∗ (𝑞 ∗ 𝑥) = 0 then 𝑥 ∈ 𝑉 for all 𝑥 ∈ 𝑋.

Then there exist a topology on 𝑋 for which is a fundamental system of open sets containing 0 and 𝑉(𝑎) is an open set for all 𝑉 ∈ 𝛺 and for all 𝑎 ∈ 𝑋. Moreover, (𝑋, 𝜏_Ω) is a TPUP-algebra.

Proof.Let 𝜏_Ω= {𝑂 ⊆ 𝑋: ∀ 𝑎 ∈ 𝑂, ∃ 𝑉 ∈ 𝛺 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑉(𝑎) ⊆ 𝑂}. First, we have to show that 𝜏_Ω is a topology on 𝑋. Clearly, 𝑋, 𝜙 ∈ 𝜏Ω. Let 𝑂𝜆 ∈ 𝜏Ω for some 𝜆 ∈ 𝑀 and let 𝑎 ∈ ⋃𝜆∈𝑀𝑂𝜆. Then 𝑎 ∈ 𝑂𝜆 for some 𝜆 ∈ 𝑀, so there

exist 𝑉 such that 𝑉(𝑎) ⊆ 𝑂_𝜆 and thus ⋃_𝜆∈𝑀𝑂_𝜆 ∈ 𝜏_Ω. Now, suppose that 𝑂₁,𝑂₂∈ 𝜏_Ω and let 𝑎 ∈ 𝑂₁∩ 𝑂₂. Thus, there exist 𝑉₁ and 𝑉₂ such that 𝑉₁(𝑎) ⊆ 𝑂₁ and 𝑉₂(𝑎) ⊆ 𝑂₂. Since 𝛺 is a base filter then there exist 𝑉 such that 𝑉 ⊆ 𝑉1∩ 𝑉2. Thus, we have 𝑉(𝑎) ⊆ (𝑉1∩ 𝑉2)(𝑎) ⊆ 𝑉1(𝑎) ∩ 𝑉2(𝑎) ⊆ 𝑂1∩ 𝑂2 and so 𝑂1∩ 𝑂2 ∈ 𝜏Ω. Then𝜏Ωis a

topology on 𝑋.

Now, we have to show that 𝛺 is a filter base of an open set containing 0 with respect to 𝜏_Ω. Since 𝑝 ∙ (𝑞 ∙ 0) = 0 and 𝑝 ∗ (𝑞 ∗ 0) = 0, for any 𝑝, 𝑞 ∈ 𝑉 then by (2) 0 ∈ 𝑉 (i.e., each element 𝑉 ∈ 𝛺 contains 0). If 𝑥 ∈ 𝑉(𝑝), then 𝑥 ∙ 𝑝, 𝑥 ∗ 𝑝,𝑝 ∙ 𝑥, 𝑝 ∗ 𝑥 ∈ 𝑉 and so 𝑣 = 𝑝 ∙ 𝑥, and 𝑣 = 𝑝 ∗ 𝑥. Hence, 𝑣 ∙ (𝑝 ∙ 𝑥) = 0 and 𝑣 ∗ (𝑝 ∗ 𝑥) = 0 implies that 𝑥 ∈ 𝑉. Thus, 𝑉(𝑝) ⊆ 𝑉 and 𝑉 ∈ 𝜏_Ω. If we suppose that 𝑉 is an open set containing 0, then there exist a 𝑈 ∈ 𝛺 such that 𝑈(0) ⊆ 𝑉. Then for some 𝑎 ∈ 𝑈 we note that 0 ∙ 𝑎, 0 ∗ 𝑎 ∈ 𝑈 and 𝑎 ∙ 0,𝑎 ∗ 0 ∈ 𝑈. Thus, 𝑎 ∈ 𝑈(0) and so 0 ∈ 𝑈 ⊆ 𝑈(0) ⊆ 𝑉. Then 𝛺 is a fundamental system of open sets containing 0 with respect to 𝜏_Ω.

Next, we have to show that 𝑉(𝑎) is an open in 𝜏_Ω. Let 𝑥 ∈ 𝑉(𝑎), then we have 𝑎 ∙ 𝑥, 𝑎 ∗ 𝑥 ∈ 𝑉 and 𝑥 ∙ 𝑎, 𝑥 ∗ 𝑎 ∈ 𝑉. Then by (1), there exists 𝑂1,𝑂2∈ 𝛺 such that 𝑂1(𝑎 ∙ 𝑥) ⊆ 𝑉, 𝑂1(𝑎 ∗ 𝑥) ⊆ 𝑉 and 𝑂2(𝑥 ∙ 𝑎) ⊆ 𝑉,𝑂2(𝑥 ∗ 𝑎) ⊆

𝑉. Since 𝛺 is a base filter then there exist 𝑊 ∈ 𝛺 such that 𝑊 ⊆ 𝑂1∩ 𝑂2. Let 𝑦 ∈ 𝑊(𝑥), then 𝑦 ∙ 𝑥, 𝑦 ∗ 𝑥 ⊆ 𝑊

and 𝑥 ∙ 𝑦, 𝑥 ∗ 𝑦 ∈ 𝑊. Since, (𝑥 ∙ 𝑦) ∗ [(𝑎 ∙ 𝑥) ∗ (𝑎 ∙ 𝑦)] = 0, and (𝑥 ∗ 𝑦) ∙ [(𝑎 ∗ 𝑥) ∙ (𝑎 ∗ 𝑦)] = 0. Also, (𝑥 ∙ 𝑎) ∗ [(𝑦 ∙ 𝑥) ∗ (𝑦 ∙ 𝑎)] = 0, and (𝑥 ∗ 𝑎) ∙ [(𝑦 ∗ 𝑥) ∙ (𝑦 ∗ 𝑎)] = 0. Then by (2), we have 𝑎 ∙ 𝑦, 𝑎 ∗ 𝑦 and 𝑦 ∙ 𝑎, 𝑦 ∗ 𝑎 are contained in 𝑉. Hence, 𝑉(𝑎) is an open set.

Finally, to show that (𝑋,𝜏_Ω)is a TPUP-algebra. Let 𝑥 and 𝑦 be two elements in 𝑋. Since each open set containing 𝑥 ∙ 𝑦 and 𝑥 ∗ 𝑦 contains 𝑉(𝑥 ∙ 𝑦) and 𝑉(𝑥 ∗ 𝑦) for 𝑉 ∈ 𝛺. It is enough to show that 𝑉(𝑥) ∙ 𝑉(𝑦) ⊆ 𝑉(𝑥 ∙ 𝑦) and 𝑉(𝑥) ∗ 𝑉(𝑦) ⊆ 𝑉(𝑥 ∗ 𝑦). Let 𝑢 ∙ 𝑣 ∈ 𝑉(𝑥) ∙ 𝑉(𝑦) and 𝑢 ∗ 𝑣 ∈ 𝑉(𝑥) ∗ 𝑉(𝑦), then 𝑢 ∈ 𝑉(𝑥) and 𝑣 ∈ 𝑉(𝑦). Therefore, 𝑢 ∙ 𝑥, 𝑢 ∗ 𝑥, 𝑥 ∙ 𝑢,𝑥 ∗ 𝑢, 𝑣 ∙ 𝑦, 𝑣 ∗ 𝑦, 𝑦 ∙ 𝑣, 𝑦 ∗ 𝑣 ∈ 𝑉 and so we have

𝑦 ∙ 𝑣 ≤ (𝑥 ∙ 𝑦) ∗ (𝑥 ∙ 𝑣) ≤ (𝑥 ∙ 𝑦) ∗ [(𝑢 ∙ 𝑥) ∗ (𝑢 ∙ 𝑣)] = (𝑢 ∙ 𝑥) ∗ [(𝑥 ∙ 𝑦) ∗ (𝑢 ∙ 𝑣)], and

𝑦 ∗ 𝑣 ≤ (𝑥 ∗ 𝑦) ∙ (𝑥 ∗ 𝑣) ≤ (𝑥 ∗ 𝑦) ∙ [(𝑢 ∗ 𝑥) ∙ (𝑢 ∗ 𝑣)] = (𝑢 ∗ 𝑥) ∙ [(𝑥 ∗ 𝑦) ∙ (𝑢 ∗ 𝑣)].

Hence, (𝑦 ∙ 𝑣) ∗ [(𝑢 ∙ 𝑥) ∗ ((𝑥 ∙ 𝑦) ∗ (𝑢 ∙ 𝑣))] = 0 and (𝑦 ∗ 𝑣) ∙ [(𝑢 ∗ 𝑥) ∙ ((𝑥 ∗ 𝑦) ∙ (𝑢 ∗ 𝑣))] = 0. Then by (2), we have (𝑥 ∙ 𝑦) ∗ (𝑢 ∙ 𝑣),(𝑥 ∗ 𝑦) ∙ (𝑢 ∗ 𝑣) ∈ 𝑉 and by the same way we can obtain that (𝑢 ∙ 𝑣) ∗ (𝑥 ∙ 𝑦),(𝑢 ∗ 𝑣) ∙ (𝑥 ∗ 𝑦) ∈ 𝑉. Therefore, 𝑢 ∙ 𝑣 ∈ 𝑉(𝑥 ∙ 𝑣) and𝑢 ∗ 𝑣 ∈ 𝑉(𝑥 ∗ 𝑦) which implies that 𝑉(𝑥) ∙ 𝑉(𝑦) ⊆ 𝑉(𝑥 ∙ 𝑦) and 𝑉(𝑥) ∗ 𝑉(𝑦) ⊆ 𝑉(𝑥 ∗ 𝑦). Hence, (𝑋,𝜏Ω) is a TPUP-algebra.

Example 5.2. Let 𝑋 = {0, 𝑎, 𝑏, 𝑐} with two binary operations ∙ and ∗ defined by the following Cayley tables:

Table 4.A pseudo-UP algebra with:x ∙ (y ∙ z) = y ∙ (x ∙ z) and x*(y*z) = y*(x*z) ∀ x,y, z ∈ X.

Example 5.3. If 𝑋 is a pseudo-UP algebra satisfying𝑥 ∙ (𝑦 ∙ 𝑧) = 𝑦 ∙ (𝑥 ∙ 𝑧), and 𝑥 ∗ (𝑦 ∗ 𝑧) = 𝑦 ∗ (𝑥 ∗ 𝑧) for all

𝑥,𝑦, 𝑧 ∈ 𝑋, then the filter base of pseudo-UP ideal ℐof 𝑋 is a base filter satisfies conditions in Proposition 5.1. Since for every 𝑥 ∈ ℐ, ℐ(𝑥) ⊆ ℐ. Hence, the condition (1) satisfies. Now, if 𝑝, 𝑞 ∈ ℐ and 𝑝 ∙ (𝑞 ∙ 𝑥) = 0 ∈ ℐ, 𝑝 ∗ (𝑞 ∗ 𝑥) = 0 ∈ ℐ then 𝑝 ∙ 𝑥 ∈ ℐ and 𝑝 ∗ 𝑥 ∈ ℐ. Again, since 𝑝 ∈ ℐ, then by Proposition 2.5𝑥 ∈ ℐ. Hence, the condition (2) satisfies. Therefore, the topology induced by ℐ is a TPUP-algebra.

Let 𝐴 be any subset of a TPUP-algebra 𝑋 whose topology is the topology generated be a filter base satisfies all conditions of Proposition 5.1, we mean that 𝑉(𝐴) = ⋃_𝑎∈𝐴𝑉(𝑎)which is clearly an open set containing 𝐴. Thus, we have the following result.

Proposition 5.4. Let 𝐴 be any subset of a TPUP-algebra 𝑋, then 𝑐𝑙(𝐴) =∩ {𝑉(𝐴): 𝑉 ∈ 𝛺}. c b a 0 ∙ c b a 0 0 c 0 0 0 a c 0 a 0 b 0 b a 0 c c b a 0 ∗ c b a 0 0 c b 0 0 a c 0 a 0 b 0 b a 0 c

6416

Proof. Let 𝑥 ∈ 𝑐𝑙(𝐴) and 𝑉 ∈ 𝛺. Since 𝑉(𝑥) is an open set containing 𝑥, then 𝑉(𝑥) ∩ 𝐴 ≠ 𝜙. Therefore, there

exist 𝑎 ∈ 𝐴 such that 𝑎 ∙ 𝑥, 𝑎 ∗ 𝑥 ∈ 𝑉 and 𝑥 ∙ 𝑎, 𝑥 ∗ 𝑎 ∈ 𝑉. Then 𝑥 ∈ 𝑉(𝑎) and 𝑥 ∈ ∩ {𝑉(𝐴): 𝑉 ∈ 𝛺}. Conversely, if 𝑥 ∈ ∩ {𝑉(𝐴): 𝑉 ∈ 𝛺}, then for any 𝑈 ∈ 𝛺we have 𝑥 ∈ 𝑈(𝐴). Therefore, 𝑈(𝑥) ∩ 𝐴 ≠ 𝜙.

Proposition 5.5. Let 𝐴 be a compact subset of a TPUP-algebra 𝑋. If 𝑈 is an open set containing 𝐴, then there exist

𝑉 ∈ 𝛺 such that 𝐴 ⊆ 𝑉(𝐴) ⊆ 𝑈.

Proof.Since 𝑈 is an open set of 𝐴, then by Proposition 5.1 for all 𝑎 ∈ 𝐴 there exist 𝑉_𝑎∈ 𝛺 such that 𝑉_𝑎(𝑎) ⊆ 𝑈.

Since, 𝐴 is a compact and 𝐴 ⊆ ⋃_𝑎∈𝐴𝑉_𝑎(𝑎) then there exist 𝑎₁,𝑎₂,. . . ,𝑎_𝑛such that 𝐴 ⊆ 𝑉_𝑎1(𝑎₁) ∪ 𝑉_𝑎2(𝑎₂) ∪ . . .∪ 𝑉𝑎𝑛(𝑎𝑛). Now, let 𝑉 = ⋃ 𝑉𝑛𝑖=1 𝑎𝑖(𝑎𝑖) so it is enough to show that 𝑉(𝑎) ⊆ 𝑈 for all 𝑎 ∈ 𝐴. Since 𝑎 ∈ 𝑉𝑎𝑖 for some

𝑎𝑖, then 𝑎 ∙ 𝑎𝑖,𝑎 ∗ 𝑎𝑖∈ 𝑉𝑎𝑖 and 𝑎𝑖∙ 𝑎,𝑎𝑖∗ 𝑎 ∈ 𝑉𝑎𝑖. If 𝑥 ∈ 𝑉(𝑎), then we have 𝑎 ∙ 𝑥, 𝑎 ∗ 𝑥 ∈ 𝑉 and 𝑥 ∙ 𝑎, 𝑥 ∗ 𝑎 ∈ 𝑉.

Since (𝑎 ∙ 𝑥) ∗ [(𝑎_𝑖∙ 𝑎) ∗ (𝑎_𝑖∙ 𝑥)] = 0, and (𝑎 ∗ 𝑥) ∙ [(𝑎_𝑖∗ 𝑎) ∙ (𝑎_𝑖∗ 𝑥)] = 0. Then by Proposition 5.1, we have 𝑎𝑖∙ 𝑥, 𝑎𝑖∗ 𝑥 ∈ 𝑉𝑎𝑖. Similarly, (𝑥 ∙ 𝑎) ∗ [(𝑎 ∙ 𝑎𝑖) ∗ (𝑥 ∙ 𝑎𝑖)] = 0 and (𝑥 ∗ 𝑎) ∙ [(𝑎 ∗ 𝑎𝑖) ∙ (𝑥 ∗ 𝑎𝑖)] = 0. Hence,

𝑥 ∙ 𝑎𝑖,𝑥 ∗ 𝑎𝑖∈ 𝑉𝑎𝑖. Therefore, 𝑥 ∈ 𝑉𝑎𝑖(𝑎𝑖) ⊆ 𝑈 and 𝑉(𝑎) ⊆ 𝑈 and thus 𝑉(𝐴) ⊆ 𝑈.

Proposition 5.6. Let 𝐾 be a compact subset of a TPUP-algebra 𝑋 and 𝐹 be a closed subset of 𝑋. If 𝐾 ∩ 𝑉 = 𝜙, so

there exist 𝑉 ∈ 𝛺 such that 𝑉(𝐾) ∩ 𝑉(𝐹) = 𝜙.

Proof. Since 𝑋\𝐹 is an open set of 𝐾, then by Proposition 5.5, there exist 𝑉 ∈ 𝛺 such that 𝑉(𝐾) ⊆ 𝑋\𝐹. Suppose

that 𝑉(𝐾) ∩ 𝑉(𝐹) ≠ 𝜙for every 𝑉 ∈ 𝛺. Then there exist 𝑥 ∈ 𝑉(𝐾) ∩ 𝑉(𝐹). Thus, 𝑥 ∈ 𝑉(𝑘) and 𝑥 ∈ 𝑉(𝑓) for some 𝑘 ∈ 𝐾 and for some 𝑓 ∈ 𝐹. Since, (𝑥 ∙ 𝑓) ∗ [(𝑘 ∙ 𝑥) ∗ (𝑘 ∙ 𝑓)] = 0, (𝑥 ∗ 𝑓) ∙ [(𝑘 ∗ 𝑥) ∙ (𝑘 ∗ 𝑓)] = 0 and (𝑥 ∙ 𝑘) ∗ [(𝑓 ∙ 𝑥) ∗ (𝑓 ∙ 𝑘)] = 0, (𝑥 ∗ 𝑘) ∙ [(𝑓 ∗ 𝑥) ∙ (𝑓 ∗ 𝑘)] = 0. Then by Proposition 5.1, we have 𝑘 ∙ 𝑓, 𝑘 ∗ 𝑓 ∈ 𝑉 and 𝑓 ∙ 𝑘, 𝑓 ∗ 𝑘 ∈ 𝑉. Therefore, 𝑓 ∈ 𝑉(𝑘) which is a contradiction. Hence, 𝑉(𝐾) ∩ 𝑉(𝐹) = 𝜙.

6. Conclusion

In this article, some properties of UP-ideals are extended to pseudo UP-ideals. Using pseudo UP-ideals we constructed a uniform structure on UP algebras. several topological properties andrelations among pseudo-UP algebras are obtained by using pseudo-pseudo-UP isomorphisms. In the last sectionwe generated a new topology from a filter base defined a pseudo-UP algebra and several results areobtained.

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